Diverietti's answer is exhaustive with respect to the first four questions, so let me say something just about the last one.

Let $n$ and $q$ be natural numbers with $0 < q < n$, $(n,q)=1$ and let $\xi_n$ be a primitive $n-$th root of unity. Let us consider the action of the cyclic group $\mathbb{Z}_n=\langle \xi_n \rangle$ on $\mathbb{C}^2$ defined by

$\xi_n \cdot (x,y)=(\xi_nx, \xi_n^qy)$.

Then the analytic space $X_{n,q}=\mathbb{C}^2 / \mathbb{Z}_n$ has a cyclic quotient singularity of type $\frac{1}{n}(1,q)$, and $X_{n,q} \cong X_{n', q'}$ if and only if $n=n'$ and either $q=q'$ or $qq' \equiv 1$ (mod $n$). The exceptional divisor on the minimal resolution $\tilde{X}_{n,q}$ of $X_{n,q}$ is a H-J string (abbreviation of Hirzebruch-Jung string), that is to say, a connected union $E=\bigcup_{i=1}^k Z_i$ of smooth rational curves $Z_1, \ldots, Z_k$ with self-intersection $\leq -2$, and ordered linearly so that $Z_i Z_{i+1}=1$ for all $i$, and $Z_iZ_j=0$ if $|i-j| \geq 2$. More precisely, given the continued fraction

$ \frac{n}{q}=[b_1,\ldots,b_k]=b_1- \cfrac{1}{b_2 -\cfrac{1}{\dotsb - \cfrac{1}{\,b_k}}}, \quad b_i\geq 2$,

the dual graph of $E$ is

${\setlength{\unitlength}{1.1cm} \begin{center} \begin{picture}(1,0.5) \put(0,0){\circle*{0.2}} \put(1,0){\circle*{0.2}} \put(0,0){\line(1,0){1}} \put(-0.3,0.2){\scriptsize $-b_1$} \put(0.70,0.2){\scriptsize $-b_2$} \put(2,0){\circle*{0.2}} \put(1,0){\line(1,0){0.2}} \put(1.3,0){\line(1,0){0.15}} \put(1.55,0){\line(1,0){0.15}} \put(1.8,0){\line(1,0){0.2}} \put(3,0){\circle*{0.2}} \put(2,0){\line(1,0){1}} \put(1.70,0.2){\scriptsize $-b_{k-1}$} \put(2.70,0.2){\scriptsize $-b_k$} \end{picture} \hspace{2.5cm} \end{center} }$

Diverietti's answer is exhaustive with respect to the first four questions, so let me say something just about the last one.

Miles Reid's example is actually a particular case of the following more general situation, which is explained for instance in Barth-Peters-Van de Ven book on compact complex surfaces.

Let $n$ and $q$ be natural numbers with $0 < q < n$, $(n,q)=1$ and let $\xi_n$ be a primitive $n-$th root of unity. Let us consider the action of the cyclic group $\mu_n=\langle \xi_n \rangle$ on $\mathbb{C}^2$ defined by

$\xi_n \cdot (x,y)=(\xi_nx, \xi_n^qy)$.

Then the analytic space $X_{n,q}=\mathbb{C}^2 / \mu_n$ has a cyclic quotient singularity of type $\frac{1}{n}(1,q)$, and $X_{n,q} \cong X_{n', q'}$ if and only if $n=n'$ and either $q=q'$ or $qq' \equiv 1$ (mod $n$). The exceptional divisor on the minimal resolution $\tilde{X}_{n,q}$ of $X_{n,q}$ is a H-J string (abbreviation of Hirzebruch-Jung string), that is to say, a connected union $E=\bigcup_{i=1}^k Z_i$ of smooth rational curves $Z_1, \ldots, Z_k$ with self-intersection $\leq -2$, and ordered linearly so that $Z_i Z_{i+1}=1$ for all $i$, and $Z_iZ_j=0$ if $|i-j| \geq 2$. More precisely, given the continued fraction

$\frac{n}{q}=[b_1,\ldots,b_k]:=b_1- \cfrac{1}{b_2 -\cfrac{1}{\dotsb - \cfrac{1}{\,b_k}}}$,

we have

$(Z_i)^2=-b_i, \quad i=1, \ldots, k.$

In the case considered by Miles Reid, we have $n=3$, $q=1$, i.e. the action is

$\xi_3 \cdot (x,y)=(\xi_3x, \xi_3y)$,

hence the resolution is a unique smooth curve $E:=Z_1$ with self-intersection $(-3)$; this explain why $\mathcal{O}_E(-E)=\mathcal{O}(3)$.

Notice that there is another possible action of $\mu_3$ on $\mathbb{C}^2$, namely

$\xi_3 \cdot (x,y)=(\xi_3x, \xi_3^2y)$.

The corresponding continued fraction is

$\frac{3}{2}=[2,2]=2- \frac{1}{2}$,

so the resolution is this case is given by two smooth rational curves of self-intersection $(-2)$ intersecting in a single point (this is a Rational Double Point of type $A_2$).